Using the concept of slope,prove that medians of an equilateral triangle are perpendicular to the corresponding sides.

Let ABC be an equilateral triangle having each side as 2a and BC be along X-axis. Also, let mid-point of BC be the origin O(0, 0) and the perpendicular to BC, which passes passing through O, be along Y-axis. Now, the coordinates of B and C are (–a, 0) and (a, 0) respectively.

Let the coordinates of A be (α,b).

It is given that DABC is an equilateral triangle, therefore

AB = AC = BC = 2a

Now,

AB = AC

⇒AB² = AC²

⇒ (a + a)² + (b – 0)² = (α – a)² + (b – 0)²

⇒ 4aα = 0

⇒ α = 0

On applying Pythagoras theorem in right triangle COA, we have

Thus, we get the coordinates of A as and point A lies on Y-axis, therefore

 OA ⊥ BC  [OA is along Y-axis and BC is along X-axis]

Let D and E be the mid-points of AC and AB respectively. Then, the coordinates of D and E will be respectively.

Similarly,

∴AB ⊥ CE

Thus, AO, BD and CE are medians of the equilateral triangle ABC in such a way that AO ⊥ BC, BD ⊥ AC and CE ⊥ AB.

Hence, medians of an equilateral triangle are perpendicular to the corresponding sides.